Question

Gribbles are small, pale white, marine worms that bore through wood. While sometimes considered a pest since they can wreck wooden docks and piers, they are now being studied to determine whether the enzyme they secrete will allow us to turn inedible wood and plant waste into biofuel.1 A sample of 50 gribbles finds an average length of 3.1 mm with a standard deviation of 0.72. Give a best estimate for the length of gribbles, a margin of error for this estimate (with 95% confidence), and a 95% confidence interval.

Answer 1

Answer:

The best estimate for the length of gribbles is of 3.1 mm.

The margin of error for this estimate, with 95% confidence, is of 0.2 mm.

The 95% confidence interval is between 2.9 mm and 3.3 mm.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

Best estimate for the length of gribbles

The best estimate is the sample mean, that is 3.1 mm.

Margin of error:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 50 - 1 = 49

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.01

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.01\frac{0.72}{\sqrt{50}} = 0.2$

In which s is the standard deviation of the sample and n is the size of the sample.

The margin of error for this estimate, with 95% confidence, is of 0.2 mm.

Confidence interval:

The lower end of the interval is the sample mean subtracted by M. So it is 3.1 mm - 0.2 mm = 2.9 mm

The upper end of the interval is the sample mean added to M. So it is 3.1 mm + 0.2 mm = 3.3 mm

The 95% confidence interval is between 2.9 mm and 3.3 mm.